Determining and justifying the validity of an argument












2












$begingroup$


Context: Question made up by uni lecturer



Original statement: There exists two positive real numbers $x$ and $y$ such that for all positive integers $z$, $frac{x}{y}>z$.



So the question was to find the negation of the statement, and then determine whether the original statement or its negation was true.



I found its negation to be: For all positive real numbers $x$ and $y$, there exists a positive integer $z$ such that $frac{x}{y}le z$.



The lecturer's solution to the question says that the negation is true since for any positive reals $x$ and $y$, you can choose $z$ to equal the ceiling of $frac{x}{y}$.



When I attempted the question myself, I said that the original statement is true because you can take $x=z+1$ (which would be a positive integer that still belongs to the set of all positive real numbers) and $y=1$ (which is a positive real number), as $frac{x}{y}=frac{z+1}{1}=z+1>z$.



Can someone please help me to see the error in my answer.



Thanks






discrete-mathematics







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$endgroup$

















    2












    $begingroup$


    Context: Question made up by uni lecturer



    Original statement: There exists two positive real numbers $x$ and $y$ such that for all positive integers $z$, $frac{x}{y}>z$.



    So the question was to find the negation of the statement, and then determine whether the original statement or its negation was true.



    I found its negation to be: For all positive real numbers $x$ and $y$, there exists a positive integer $z$ such that $frac{x}{y}le z$.



    The lecturer's solution to the question says that the negation is true since for any positive reals $x$ and $y$, you can choose $z$ to equal the ceiling of $frac{x}{y}$.



    When I attempted the question myself, I said that the original statement is true because you can take $x=z+1$ (which would be a positive integer that still belongs to the set of all positive real numbers) and $y=1$ (which is a positive real number), as $frac{x}{y}=frac{z+1}{1}=z+1>z$.



    Can someone please help me to see the error in my answer.



    Thanks






    discrete-mathematics







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Context: Question made up by uni lecturer



      Original statement: There exists two positive real numbers $x$ and $y$ such that for all positive integers $z$, $frac{x}{y}>z$.



      So the question was to find the negation of the statement, and then determine whether the original statement or its negation was true.



      I found its negation to be: For all positive real numbers $x$ and $y$, there exists a positive integer $z$ such that $frac{x}{y}le z$.



      The lecturer's solution to the question says that the negation is true since for any positive reals $x$ and $y$, you can choose $z$ to equal the ceiling of $frac{x}{y}$.



      When I attempted the question myself, I said that the original statement is true because you can take $x=z+1$ (which would be a positive integer that still belongs to the set of all positive real numbers) and $y=1$ (which is a positive real number), as $frac{x}{y}=frac{z+1}{1}=z+1>z$.



      Can someone please help me to see the error in my answer.



      Thanks






      discrete-mathematics







      share|cite|improve this question









      $endgroup$




      Context: Question made up by uni lecturer



      Original statement: There exists two positive real numbers $x$ and $y$ such that for all positive integers $z$, $frac{x}{y}>z$.



      So the question was to find the negation of the statement, and then determine whether the original statement or its negation was true.



      I found its negation to be: For all positive real numbers $x$ and $y$, there exists a positive integer $z$ such that $frac{x}{y}le z$.



      The lecturer's solution to the question says that the negation is true since for any positive reals $x$ and $y$, you can choose $z$ to equal the ceiling of $frac{x}{y}$.



      When I attempted the question myself, I said that the original statement is true because you can take $x=z+1$ (which would be a positive integer that still belongs to the set of all positive real numbers) and $y=1$ (which is a positive real number), as $frac{x}{y}=frac{z+1}{1}=z+1>z$.



      Can someone please help me to see the error in my answer.



      Thanks






      discrete-mathematics




      discrete-mathematics






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked Mar 26 at 13:14









      Ruby PaRuby Pa

      376




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          2 Answers
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          $begingroup$

          The problem with your reasoning is the order. There exist $x,y$ positive integers such that for all positive integers $z$ we have $frac{x}{y}>z$, so first you must pick an $x$ and a $y$, and then you must test whether $frac{x}{y}>z$ for every positive integer $z$. Therefore you can't define $x=z+1$ as when you pick $x$ you don't know $z$ yet.






          share|cite|improve this answer









          $endgroup$





















            3












            $begingroup$


            "There exists two positive real numbers $x$ and $y$ such that for all positive integer $z$ : $dfrac x y > z$"




            The intuition says that it is False. $dfrac x y$ is a positive real; thus, the statement amounts to asserting that there is a real that is greater than every integer, which is not.



            You reasoning is wrong because you have swapped the choice of the numbers : you start from $z$ and choose $x$ and $y$ accordingly.



            The negation of the original statement is : $forall x forall y exists z (dfrac y y le z)$.



            Thus, choose $x$ and $y$ positive whatever and what you get is a new positive real $dfrac x y$.



            Now, you have to choose an integer $z$ (obviously positive) that is greater-or-equal to $dfrac x y$.



            And this must be always possible, because $dfrac x y$ is a number $r.r_1 r_2 r_3 ldots$.



            Consider as $z$ the number $r+1$ and it's done.






            share|cite|improve this answer











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              2 Answers
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              2 Answers
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              $begingroup$

              The problem with your reasoning is the order. There exist $x,y$ positive integers such that for all positive integers $z$ we have $frac{x}{y}>z$, so first you must pick an $x$ and a $y$, and then you must test whether $frac{x}{y}>z$ for every positive integer $z$. Therefore you can't define $x=z+1$ as when you pick $x$ you don't know $z$ yet.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                The problem with your reasoning is the order. There exist $x,y$ positive integers such that for all positive integers $z$ we have $frac{x}{y}>z$, so first you must pick an $x$ and a $y$, and then you must test whether $frac{x}{y}>z$ for every positive integer $z$. Therefore you can't define $x=z+1$ as when you pick $x$ you don't know $z$ yet.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  The problem with your reasoning is the order. There exist $x,y$ positive integers such that for all positive integers $z$ we have $frac{x}{y}>z$, so first you must pick an $x$ and a $y$, and then you must test whether $frac{x}{y}>z$ for every positive integer $z$. Therefore you can't define $x=z+1$ as when you pick $x$ you don't know $z$ yet.






                  share|cite|improve this answer









                  $endgroup$



                  The problem with your reasoning is the order. There exist $x,y$ positive integers such that for all positive integers $z$ we have $frac{x}{y}>z$, so first you must pick an $x$ and a $y$, and then you must test whether $frac{x}{y}>z$ for every positive integer $z$. Therefore you can't define $x=z+1$ as when you pick $x$ you don't know $z$ yet.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 26 at 13:33









                  Floris ClaassensFloris Claassens

                  1,33229




                  1,33229























                      3












                      $begingroup$


                      "There exists two positive real numbers $x$ and $y$ such that for all positive integer $z$ : $dfrac x y > z$"




                      The intuition says that it is False. $dfrac x y$ is a positive real; thus, the statement amounts to asserting that there is a real that is greater than every integer, which is not.



                      You reasoning is wrong because you have swapped the choice of the numbers : you start from $z$ and choose $x$ and $y$ accordingly.



                      The negation of the original statement is : $forall x forall y exists z (dfrac y y le z)$.



                      Thus, choose $x$ and $y$ positive whatever and what you get is a new positive real $dfrac x y$.



                      Now, you have to choose an integer $z$ (obviously positive) that is greater-or-equal to $dfrac x y$.



                      And this must be always possible, because $dfrac x y$ is a number $r.r_1 r_2 r_3 ldots$.



                      Consider as $z$ the number $r+1$ and it's done.






                      share|cite|improve this answer











                      $endgroup$


















                        3












                        $begingroup$


                        "There exists two positive real numbers $x$ and $y$ such that for all positive integer $z$ : $dfrac x y > z$"




                        The intuition says that it is False. $dfrac x y$ is a positive real; thus, the statement amounts to asserting that there is a real that is greater than every integer, which is not.



                        You reasoning is wrong because you have swapped the choice of the numbers : you start from $z$ and choose $x$ and $y$ accordingly.



                        The negation of the original statement is : $forall x forall y exists z (dfrac y y le z)$.



                        Thus, choose $x$ and $y$ positive whatever and what you get is a new positive real $dfrac x y$.



                        Now, you have to choose an integer $z$ (obviously positive) that is greater-or-equal to $dfrac x y$.



                        And this must be always possible, because $dfrac x y$ is a number $r.r_1 r_2 r_3 ldots$.



                        Consider as $z$ the number $r+1$ and it's done.






                        share|cite|improve this answer











                        $endgroup$
















                          3












                          3








                          3





                          $begingroup$


                          "There exists two positive real numbers $x$ and $y$ such that for all positive integer $z$ : $dfrac x y > z$"




                          The intuition says that it is False. $dfrac x y$ is a positive real; thus, the statement amounts to asserting that there is a real that is greater than every integer, which is not.



                          You reasoning is wrong because you have swapped the choice of the numbers : you start from $z$ and choose $x$ and $y$ accordingly.



                          The negation of the original statement is : $forall x forall y exists z (dfrac y y le z)$.



                          Thus, choose $x$ and $y$ positive whatever and what you get is a new positive real $dfrac x y$.



                          Now, you have to choose an integer $z$ (obviously positive) that is greater-or-equal to $dfrac x y$.



                          And this must be always possible, because $dfrac x y$ is a number $r.r_1 r_2 r_3 ldots$.



                          Consider as $z$ the number $r+1$ and it's done.






                          share|cite|improve this answer











                          $endgroup$




                          "There exists two positive real numbers $x$ and $y$ such that for all positive integer $z$ : $dfrac x y > z$"




                          The intuition says that it is False. $dfrac x y$ is a positive real; thus, the statement amounts to asserting that there is a real that is greater than every integer, which is not.



                          You reasoning is wrong because you have swapped the choice of the numbers : you start from $z$ and choose $x$ and $y$ accordingly.



                          The negation of the original statement is : $forall x forall y exists z (dfrac y y le z)$.



                          Thus, choose $x$ and $y$ positive whatever and what you get is a new positive real $dfrac x y$.



                          Now, you have to choose an integer $z$ (obviously positive) that is greater-or-equal to $dfrac x y$.



                          And this must be always possible, because $dfrac x y$ is a number $r.r_1 r_2 r_3 ldots$.



                          Consider as $z$ the number $r+1$ and it's done.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Mar 26 at 13:44

























                          answered Mar 26 at 13:29









                          Mauro ALLEGRANZAMauro ALLEGRANZA

                          67.7k449117




                          67.7k449117






























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